STD 12 - 5. continuity and differentiation — JEE कक्षा 11 साइन्स — Question

$x$  के सापेक्ष अवकलन करने पर,

$1 = \frac{{dy}}{{dx}}\sqrt {1 - {y^2}} + y.\frac{1}{{2\sqrt {1 - {y^2}} }}\,.\,( - 2y)\,.\,\frac{{dy}}{{dx}}$

==> $1 = \frac{{dy}}{{dx}}\sqrt {1 - {y^2}} - \frac{{{y^2}}}{{\sqrt {1 - {y^2}} }}\,.\,\frac{{dy}}{{dx}}$

==> $1 = \frac{{dy}}{{dx}}\left[ {\frac{{1 - {y^2} - {y^2}}}{{\sqrt {1 - {y^2}} }}} \right]$

==> $1 = \frac{{dy}}{{dx}}\left[ {\frac{{1 - 2{y^2}}}{{\sqrt {1 - {y^2}} }}} \right]$

$\frac{{dy}}{{dx}} = \frac{{\sqrt {1 - {y^2}} }}{{1 - 2{y^2}}}$.